Video Transcript
Find 𝑚 such that 40𝑃𝑚 plus 15
equals 11𝑃𝑚 plus 15.
The notation used in this equation
is permutation notation. And in general, for positive
integer 𝑛 and integer 𝑟 between zero and 𝑛 inclusive, the permutation symbol
𝑛𝑃𝑟 represents the integer given by the calculation 𝑛 factorial divided by 𝑛
minus 𝑟 factorial. We call it a permutation symbol
because this number is precisely the number of permutations of 𝑟 unique objects
selected from a collection of 𝑛 unique objects.
The notation 𝑛 factorial
represents the product of all of the positive integers from one to 𝑛 inclusive. And we can write this recursively
as 𝑛 times 𝑛 minus one factorial. This makes sense because 𝑛 minus
one factorial is the product of all of the integers from one to 𝑛 minus one. And multiplying by 𝑛 then gives us
the product of all the integers from one to 𝑛.
Okay, now we’re going to use these
definitions to tackle our equation. For starters, note that the 𝑟 of
𝑛𝑃𝑟 in both of these symbols is the same thing, 𝑚 plus 15. So to simplify our expressions and
our algebra, let’s to define another variable 𝑘 equal to 𝑚 plus 15. This turns our equation into 40𝑃𝑘
equals 11𝑃𝑘. Then, after solving this equation
for 𝑘, we simply subtract 15 to find 𝑚. It’s important to understand that
this variable substitution doesn’t change the problem. It merely makes the problem easier
to look at, which is very hopeful in trying to see a way forward.
Expanding using our definition of
𝑛𝑃𝑟, we have that 40 factorial divided by 40 minus 𝑘 factorial is equal to 11
factorial divided by 11 minus 𝑘 factorial. Remember that the definition of
factorial explicitly includes the multiplication of many terms. Also, for two numbers to be equal
to each other, they must have exactly the same prime factors. So since both sides of our equation
are a product of integers, our strategy will be to match the prime factors of both
sides. Before we start implementing this
strategy though, we have to make sure that our implicit assumptions actually
hold.
First, we stated that both sides of
the equation are products of integers. But as written, they’re, in fact,
both fractions. So what we need to do is show that
both of these fractions are equivalent to an integer or a product of integers. We’ll do this by applying the
definition of 𝑛 factorial to the definition of 𝑛𝑃𝑟. We start with 𝑛𝑃𝑟 equals 𝑛
factorial divided by 𝑛 minus 𝑟 factorial. We now use our definition to
rewrite 𝑛 factorial as 𝑛 times 𝑛 minus one factorial. But we can now apply the definition
again to rewrite 𝑛 minus one factorial as 𝑛 minus one times 𝑛 minus two
factorial.
And we’ll keep applying the
definition a total of 𝑟 times. This will give us 𝑛 times 𝑛 minus
one times 𝑛 minus two, et cetera, times 𝑛 minus 𝑟 plus one times 𝑛 minus 𝑟
factorial all divided by 𝑛 minus 𝑟 factorial. But now there is a common factor of
𝑛 minus 𝑟 factorial in the numerator and denominator. So that division just leaves us
with one. And 𝑛𝑃𝑟 is equal to 𝑛 times 𝑛
minus one times 𝑛 minus two et cetera times 𝑛 minus 𝑟 plus one. In fact, writing 𝑛𝑃𝑟 as a
product of these 𝑟 integers is often a useful way to express 𝑛𝑃𝑟 instead of
using factorials.
Regardless, we see that 𝑛𝑃𝑟 is
always an integer, which means that our strategy is indeed justified. There is, however, one more subtle
point which needs to be addressed, which is that our strategy implicitly assumes
that both sides are the product of prime factors. Although almost every positive
integer is in fact the product of prime factors, there is a single exception. The number one is not a prime
number but is also not the product of prime numbers either. So we’ll have to amend our strategy
with the caveat that this equality holds if we can match all of the prime factors of
both sides or both sides are equal to one.
Is this possible? Can our equality hold because both
sides are equal to one? Indeed, it is possible. Take a look at the left-hand
side. 40 factorial divided by 40 minus 𝑘
factorial will be one if 40 factorial is equal to 40 minus 𝑘 factorial. But this equality will hold if and
only if 40 is equal to 40 minus 𝑘. And of course, if 40 equals 40
minus 𝑘, then 𝑘 equals zero.
Looking at the left-hand side, 11
factorial divided by 11 minus 𝑘 factorial with 𝑘 equals zero is just 11 factorial
divided by 11 factorial, which is also one. So the condition that both sides
are equal to one is met if 𝑘 is zero. In fact, it is generally true that
𝑛𝑃 zero equals one for all positive 𝑛. This is because 𝑛 factorial
divided by 𝑛 minus zero factorial is just 𝑛 factorial divided by 𝑛 factorial,
which is one.
Note also that this reveals that
our other expression for 𝑛𝑃𝑟 implicitly assumes that 𝑟 is greater than zero
because it doesn’t make any sense to have zero terms on the right-hand side of this
equation. And indeed, plugging in 𝑟 equals
zero definitely does not give us 𝑛𝑃 zero equals one. It’s also worth clarifying that in
representing this product we’ve shown two explicit terms because 𝑟 is
undetermined. But if 𝑟 is one, there’s actually
only one term, and that’s 𝑛.
Anyway, we’ve now completely dealt
with the special case that both sides of our equation evaluate to one. And indeed, we’ve found a solution
for this special case, namely, 𝑘 equals zero. We’ll tuck this solution away for
later and now apply our main strategy of matching the prime factors of both
sides. Remember that to do this, it’s
helpful to have both sides explicitly written as a product of factors. So to do this, we’ll want to use
this other expression for 𝑛𝑃𝑟. But as we saw, this expression is
only valid if 𝑟 is greater than zero.
To take care of this, we will do
all of the following calculations under the assumption that 𝑘 is greater than
zero. Now, since this assumption cuts out
a valid possibility that 𝑘 is equal to zero, we do have to check that 𝑘 equals
zero does or doesn’t give us a solution. Luckily, we already know that 𝑘
equals zero does give us a solution. So we’re ready to plow straight
ahead with our strategy.
Expanding out our equation then
gives us that 40 times 39, et cetera, times 40 minus 𝑘 plus one equals 11 times 10,
et cetera, times 11 minus 𝑘 plus one. We’ve put this little question mark
around the equals sign because we don’t know if these two quantities are equal. That’s what we’re trying to figure
out. Also, just to be clear, the factors
of 39 and 10 are only there if 𝑘 is two or greater. We’ve written them out just to see
what’s happening. But we also understand that if we
discover that 𝑘 is less than two, then we have to remove these factors from the
equation.
Anyway, let’s get started with
matching those prime factors. Since 𝑘 is greater than zero by
assumption, the right-hand side will always include the prime factor 11. Furthermore, since 11 is the
largest number on the right-hand side, no matter what the value of 𝑘 is, the
right-hand side will never be divisible by any prime number greater than 11. Let’s now compare each of these two
conditions to the left-hand side.
Because 11 is prime, for there to
be a factor of 11 on the left-hand side, one of the terms in this product must be
divisible by 11. Now observe that all of the terms
on the left-hand side are less than or equal to 40 and 40 is not divisible by
11. The multiple of 11 closest to but
not exceeding 40 is 33, which is three times 11. So for the left-hand side to be
divisible by 11, one of the terms must be 33. To see why 33 must necessarily be a
term and not, say, 22 or 11, we observe that all the terms on the left-hand side
form a sequence of decreasing integers, which means that if 22 appears, 33 must have
already appeared 11 numbers earlier. Likewise, if 11 appears, then 33
must have already appeared 22 numbers earlier.
Anyway, before going further with
this, let’s return to the observation that no prime greater than 11 is a factor of
the right-hand side. But let’s now look at what we’ve
written for the left-hand side. One of the possible terms on the
left-hand side is 39, but 39 is three times 13. But this means that if 39 really is
a term in the product on the left-hand side, then 13 is a prime factor of this
product, which contradicts our requirement that there’d be no primes greater than
11. So we conclude that 39 cannot be a
term on the left-hand side.
But now we’ve set up a
contradiction. For all 𝑘 greater than zero, the
right-hand side will always have a prime factor of 11 and no prime factors greater
than 11. So in order for these products to
be equal, that is, in order for us to match prime factors, the left-hand side must
have a prime factor of 11 and no prime factors greater than 11. And as we saw, this means that 33
must be a term on the left-hand side and that 39 must not be a term on the left-hand
side.
But again, as we saw, the left-hand
side is a sequence of decreasing consecutive integers. This means that if 33 is really a
term on the left-hand side, then 39 must be a term because to get to 33, we had to
go 40, 39, 38, 37, 36, 35, 34, 33. But it is of course impossible to
have 39 as both a term and not a term. So our necessary conditions for the
right-hand side lead to a contradiction on the left-hand side.
But since we’re led to a
contradiction, it must be that our assumption was invalid. And our assumption was that this
equation had solutions for some 𝑘 greater than zero. So the conclusion we are forced to
reach then is that this equation has no solutions for 𝑘 greater than zero and the
only solution is the one we already found, 𝑘 equals zero.
Now all that’s left to do is use 𝑘
equals zero to solve for 𝑚. So we have zero equals 𝑚 plus 15
or 𝑚 equals negative 15 as the only solution.
